First slide

Means - A.M/G.M/H.M

Question

$If x, y, z \in ℝ^{+} such that x + y + z = 4, then maximum possible value of x y z^{2} is:$

Moderate

A

1
B

2
C

3
D

4

Solution

$Given that x + y + z = 4$
$\frac{x + y + z}{4} = 1$
$Arithmetic mean (A.M) \geq Geometric mean(G.M)$
$\begin{array}{l} \frac{x + y + \frac{z}{2} + \frac{z}{2}}{4} \geq {(x . y . \frac{z}{2} . \frac{z}{2})}^{\frac{1}{4}} \\ \Rightarrow 1 \geq {(\frac{x y z^{2}}{4})}^{\frac{1}{4}} \end{array}$
$\Rightarrow x y z^{2} \leq 4$
$Maximum value of x y z^{2} is 4$

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